I am an amateur and a beginner in the topic

A theorem states

Let K be a field of characteristic p

an elliptic curve is supersingular iff $card(E(K)) = 1$ mod $ p$

supersingular means: the subgroup E[p] of points of order p of E is trivial, equal to the infinite point only (while for ordinary curves, E[p] is the cyclical group of order p)

Then lets consider this example: the curve E defined by $y^2 = x^3+2x+1$ on $F_5$

i can easily check on SageMath that its points are [(0 : 1 : 0), (0 : 1 : 1), (0 : 4 : 1), (1 : 2 : 1), (1 : 3 : 1), (3 : 2 : 1), (3 : 3 : 1)] E has the infinite point and 6 points of order 7, the group E is the cyclical group of order 7.

There are thus no points of order 5 except from the infinity point, and E[5] is trivial, which means E is supersingular

But that exemple seems to contradict the theorem, because card(E) = 7 is NOT equal to 1 mod 5.

Please could you tell me what is wrong ? Is it my understanding of the definitions ? or some bad calculation in this example ?

Thank you

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    $\begingroup$ Points on an elliptic curve in characteristic $p$ don't only have coordinates in $\mathbf F_p$. You did not tell us what definition you are using for supersingular elliptic curves in characteristic $p$. Don't confuse, for example, a property about points in the algebraic closure of $\mathbf F_p$ with a property of points in $\mathbf F_p$. $\endgroup$ – KCd Jun 1 at 7:08
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    $\begingroup$ What is true is that an elliptic curve over a finite field of characteristic $p$ is supersingular iff $|E(K)|\equiv1\pmod p$. Anyway, over the algebraic closure of $\Bbb F_5$ your curve has $|E[5]|=25$, $\endgroup$ – Angina Seng Jun 1 at 7:22
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    $\begingroup$ For a finite field $\mathbb{F_q}$, with $q=p^k$ ($p$ a prime), an elliptic curve $E$ is supersingular if and only if $E[p](\mathbb{F}_{q^r})$ is trivial for all integers $r\geq 1$. This is a particular case of the general definition for a field $K$, where $E[p](\bar{K})$ is required to be trivial. I suggest you try computing $|E(\mathbb{F}_{5^r}|$ for a few more values of $r$, and eventually it will be divisible by 5. $\endgroup$ – tkf Jun 1 at 8:02

thanks, i realize i hadn't fully absorbed the fact that E[n] is a subgroup of $E(\bar{K})$ and not of E(K).

in my particular example, there are no elements of order 5 in $E(F_5)$ but i could see effectively on Sage that $E(F_{5^{2n}})$ has 4 points of order 5 (plus the point at infinity whose order == 1 divides 5)

The theorem states that effectively E[5] is the cyclic group of order 5.

Thank you very much

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